Is there a "geometric reason" for why the gradient of $f(x,y)=\frac{x}{\sqrt{x^2+y^2}}$ is tangent to the unit circle?

Question

Consider $f(x,y)=\frac{x}{\sqrt{x^2+y^2}}$ defined on the unit $2$-dimensional disk without the origin. A direct computation shows that $\nabla f(x,y)\perp (x,y)$, so $\nabla f(x,y)$ is tangent to the circle of radius $\sqrt{{x^2+y^2}}$ at the point $(x,y)$.

Is this fact "obvious", when looking with the right glasses?

I wonder it there is a way to "see this" with as little computation as possible. One intuition I do have is that taking an "infinitesimal step" on the circle keeps the denominator $\sqrt{{x^2+y^2}}$ constant while increasing the numerator. However, this does not seem to me a rigorous explanation, since it is not clear on advance why the maximal increase is obtained at this situation (there might be a better trade-off increasing the numerator and the denominator simultaneously).

Is there a simple (rigorous) explanation?

Here are two different explanations I have (I still wish for a justification which will make it seem more "obvious"):

The first is to use the formula for the gradient in polar coordinates. Since $f$ depends only on $\theta$, the result follows.

Edit: I now think that the fact $f$ depends only on $\theta$ provides sufficient intuition for the result, even without knowing the formula for the gradient. The point is that, since $f$ depends only on $\theta$, it would be "inefficient" to change the radius as well; the best we can do is to choose the right angular direction-i.e. increasing or decreasing $\theta$.

The second is to recall that the gradient of $f$ at $(x_0,y_0)$ is normal to the level set of $f$ through $(x_0,y_0)$, i.e. normal to $\{ (x,y) \, | \, f(x,y) =f(x_0,y_0) \}$ at $(x_0,y_0)$. Geometrically, this level set corresponds to $\cos \theta=\text{const}$ in polar coordinates, which is equivalent (locally around $(x_0,y_0)$ ) to $\theta=\text{const}$, i.e. the level set is just a ray passing through the origin, and it is visually clear that its tangent line at a point $(x_0,y_0)$ is itself- $\text{span}((x_0,y_0))$.

Answer 1

I think you have basically answered the question yourself (I would argue that the fact that $f$ depends only on $\theta$ in the polar coordinates is the right intuition).

Another way to understand the result is the fact that $f(x,y)$ is homogeneous of degree $0$, that is to say, we have $$f(\lambda x, \lambda y) = f(x,y) \tag{1}$$ for $\lambda>0$. Now taking the derivative of (1) with respect to $\lambda$, we obtain $$ x \frac{\partial f}{\partial x}(\lambda x, \lambda y) +y \frac{\partial f}{\partial y}(\lambda x, \lambda y) =0$$ and setting $\lambda =1$, the result follows.

Answer 2

Consider a point $$P=(a,b)\neq (0,0)$$ inside the unit circle and call $$ k = \frac{a} {{\sqrt {a^2 + b^2 } }} $$ The k-level curve of your function is a couple of line given by factoring $$ \left( {1 - k^2 } \right)x^2 - k^2 y^2 = 0 $$ One of them is the line through the origin and P. Since the gradient is always orthogonal to the level line it is tangent to the circle through P with center at origin.

Answer 3

In polar coordinates, the function is $\cos\theta$, i.e. it does not depend on $\rho$. Thus iso-curves $f=\text{Cst}$ are straight lines from the origin and the gradient is orthogonal to them.

Answer 4

I now have two answers that satisfies me:

1. Since $f$ depends only on $\theta$, it would be "inefficient" to change the radius as well; the best we can do is to choose the right angular direction-i.e. increasing or decreasing $\theta$.

More formally, we have the following claim (which holds in more general Riemannian settings as well):

Let $x,y$ be coordinates on a $2$D-manifold, and suppose $f(x,y)=h(x)$ is a smooth function which depends only on the $x$-coordinate. Then $\nabla f \perp \frac{\partial}{\partial y}$.

Indeed, $0=df(\frac{\partial}{\partial y})=\langle \nabla f , \frac{\partial}{\partial y} \rangle$.

The same argument shows that the gradient of a function which depends on a subset of the coordinates, lies in the span of the coordinate vector-fields associated with this subset.

(In our case $x \to \theta,y \to r$).

2. The gradient of $f$ at $(x_0,y_0)$ is normal to the level set of $f$ through $(x_0,y_0)$, which in our case is a ray emanating from the origin and passing through $(x_0,y_0)$. This ray coincides with its tangent space.